Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 27J (Applied Probability)

For independent identically distributed nonnegative random variables ${\xi }_1,{\xi }_2,\dots$, define the renewal epochsThe associated renewal process isthe number of renewals by time $t$.

If $0<\mu =\mathbb{E}{\xi }_1<\infty$, the size-biased distribution of ${\xi }_1$ is the law of $\hat{\xi }$ defined byfor bounded measurable $h$. Equivalently,

The residual lifetime process, or excess process, isIf the inter-renewal distribution is non-arithmetic and $0<\mu =\mathbb{E}{\xi }_1<\infty$, then the renewal excess limit theorem saysEquivalently, $E_{\infty }$ has the law of $U\hat{\xi }$, where $U$ is uniform on $(0,1)$ and independent of the size-biased inter-renewal time $\hat{\xi }$.

Let the Poisson process $X$ have rate $\lambda$, let $\overline{F}(t)=\mathbb{P}({\xi }_1>t)$ be the survival function of an inter-renewal time of $N$, and let $\overline{H}$ be the survival function of the first event time of $Y$. Independence givesPut

At a large time, the excess of $Y$ is the minimum of the excesses of $X$ and $N$. The Poisson excess is an independent rate-$\lambda$ exponential random variable. Applying the renewal excess limit theorem to $N$ and to the assumed renewal process $Y$ givesSubstituting $\overline{F}(y)=e^{\lambda y}\overline{H}(y)$ yields the integral equation

Set $A(x)={\int }_x^{\infty }\overline{H}(y)dy$ and $c={\mu }_H/{\mu }_F<1$. Differentiating the equation almost everywhere givesSince $A^{\prime }=-\overline{H}$, it follows thatUsing $\overline{H}(0)=1$ shows that $\overline{H}(x)=e^{-x/{\mu }_H}$. Thus the first event time of $Y$ is exponential. This is the exponential first interarrival in a renewal superposition containing a Poisson process theorem.